Important Formulae for Photoelectric Effect
Category : JEE Main & Advanced
(1) \[h\nu =h{{\nu }_{0}}+{{K}_{\max }}\] and \[{{K}_{\max }}=e{{V}_{0}}\]
(2) \[{{K}_{\max }}=e{{V}_{0}}=h(\nu -{{\nu }_{0}})\]\[\Rightarrow \]\[\frac{1}{2}mv_{\max }^{2}=h(\nu -{{\nu }_{0}})\]
(3) \[{{v}_{\max }}=\sqrt{\frac{2h(\nu -{{\nu }_{0}})}{m}}\]
(4) \[{{K}_{\max }}=\frac{1}{2}mv_{\max }^{2}=e{{V}_{0}}=hc\,\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)=hc\,\left( \frac{{{\lambda }_{0}}-\lambda }{\lambda {{\lambda }_{0}}} \right)\]
(5) \[{{v}_{\max }}=\sqrt{\frac{2hc}{m}\frac{\left( {{\lambda }_{0}}-\lambda \right)}{\lambda {{\lambda }_{0}}}}\]
(6) \[{{V}_{0}}=\frac{h}{e}(\nu -{{\nu }_{0}})=\frac{hc}{e}\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)=12375\,\left( \frac{1}{\lambda }-\frac{1}{{{\lambda }_{0}}} \right)\]
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